Head to www.savemyexams.com for more awesome resources
Cambridge (CIE) IGCSE Maths:
Extended
Conditional Probability
Contents
Conditional Probability
Combined Conditional Probabilities
Page 1 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Your notes
Head to www.savemyexams.com for more awesome resources
Conditional Probability
Your notes
Conditional Probability
What is a conditional probability?
A conditional probability is the probability of something happening (A) given that something else has
already happened (B)
Examiner Tips and Tricks
Conditional probability can be written using formal notation.
|
The notation P A B describes the probability of event A occurring, given that event B has
already occurred.
(
)
Note that you are not expected to know this notation for the exam, but you are expected to be able
to calculate conditional probabilities using Venn diagrams, tree diagrams or tables.
How do I calculate conditional probabilities?
Conditional probabilities must be out of a smaller restricted set of outcomes (not out of all possible
events)
For example, if a computer randomly selects a digit from 1, 2, 3, 4, 5, 6, 7, 8, 9
P(it selects a multiple of three) =
3
9
This is not a conditional probability
There are 3 possibilities (3, 6, 9) out of all 9 possibilities
However, if you program the computer to only select from even numbers, then
P(it selects a multiple of three given that it selects an even number) =
1
4
1 possibility (6) out of only 4 possibilities (2, 4, 6, 8)
This is a conditional probability
Page 2 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
Your notes
Examiner Tips and Tricks
Look out for conditional probability questions within larger questions on two-way tables, Venn
diagrams or tree diagrams
They often use the phrase given that
Worked Example
A Venn diagram is shown below.
Find the probability that event A will occur given that event B has already occurred. Simplify your
answer.
Given that B has already happened, your probability will be out of the total number in B
Find the total number in B
14 + 28 + 7 + 35 + 21 = 105
Page 3 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
Out of this total number in B, find how many are in A
14 + 28 = 42
Your notes
Form a probability by dividing 42 by 105
42
105
Simplify your answer
P event A occurs given that event B has already occured =
(
)
2
5
Page 4 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
Combined Conditional Probabilities
Combined Conditional Probabilities
What is a combined conditional probability?
This is when you have two (or more) successive events, one after the other, and the second event
depends on (is conditional on) the first
How do I calculate combined conditional probabilities?
You need to adjust the number of outcomes as you go along
For example, selecting two cards from a pack of 52 playing cards without replacing the first card:
P(red 1st card) is 26 reds out of 52 cards
If the 1st card is not replaced, there are only 25 reds left out the remaining 51 cards
P(red 2nd card) is 25 reds out of 51 cards
P(red then red) =
26 25
×
52 51
Examiner Tips and Tricks
If a question says "two cards are drawn" then you may assume that they draw 1 card followed by
another card without replacement (the maths is the same).
Can I a tree diagram for combined conditional probabilities?
Yes, a tree diagram is a useful way to show combined conditional probabilities
For example, two counters are drawn at random from a bag of 3 blue and 8 red counters without
replacement
The probabilities are shown below
Page 5 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Your notes
Head to www.savemyexams.com for more awesome resources
Your notes
What if there are multiple possibilities within one question?
You may need a listing strategy (e.g. AAB, ABA, BAA)
You will need the or rule for multiple possibilities
P(AB or BA or AA or...) = P(AB) + P(BA) + P(AA) +...
Add the cases together
Remember that AB and BA are not the same
AB means A happened first, then B
BA means B happened first, then A
Page 6 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
Your notes
Examiner Tips and Tricks
Try not to simplify your probabilities too early as it is easier to add probabilities together when they
all have the same denominator!
Worked Example
A bag contains 10 yellow beads, 6 blue beads and 4 green beads.
A bead is taken at random from the bag and not replaced.
A second bead is then taken at random from the bag.
(a) Find the probability that both beads are different colours.
Let Y, B and G represent choosing a yellow, blue and green bead
Method 1
The probability of the beads being different colours is equal to 1 subtract the probability that the
beads are the same colour
Find the probability of both beads being the same colour
P(same colours) = P(YY) + P(BB) + P(GG)
Calculate each conditional probability separately, remembering the number of beads changes after
one is drawn and not replaced
For example, P(YY) =
10
9
×
20 19
10
9
6
5
4
3
×
+
×
+
×
20 19 20 19 20 19
Multiply the pairs of fractions together and add their results
232
380
Subtract this from 1
Page 7 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
1−
132 248
=
380 380
Your notes
Simplify the answer
62
95
Method 2
List all the possibilities of different colours
Remember that YB (yellow first, then blue) is different to BY (blue first, then yellow)
YB, BY, YG, GY, BG, GB
Use the "or" rule to add the cases together
P(different colours) = P(YB) + P(BY) + P(YG) + P(GY) + P(BG) + P(GB)
Calculate each conditional probability separately, remembering the number of beads changes after
one is drawn and not replaced
For example, P(YB) =
10
6
×
20 19
10
6
6
10 10
4
4
10
6
4
4
6
×
+
×
+
×
+
×
+
×
+
×
20 19 20 19 20 19 20 19 20 19 20 19
Multiply the pairs of fractions together and add their results
248
380
Simplify the answer
62
95
The second bead is not replaced and a third bead is taken at random from the bag.
(b) Find the probability that all three beads are the same colour.
List the possibilities
YYY, BBB, GGG
Page 8 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers
Head to www.savemyexams.com for more awesome resources
Use the "or" rule to add between cases
P(all the same colour) = P(YYY) + P(BBB) + P(GGG)
Your notes
Use conditional probabilities in each separate case, remembering the number of beads changes
after each one is drawn and not replaced
10
9
8
6
5
4
4
3
2
×
×
+
×
×
+
×
×
20 19 18 20 19 18 20 19 18
Multiply the triplets of fractions together then add their results
720
120
24
864
+
+
=
6840 6840 6840 6840
Simplify the answer
12
95
Page 9 of 9
© 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers